bonanova 78 Report post Posted January 21, 2013 It is a beginner-level bit of mathematical magic to prove that 1=2. This demonstration probably occurs multiple times in the history of this forum.Its fallacy depends on embedding, and cleverly hiding somewhere in the "analysis," a division by zero.Which of course is not permitted. So regarding this matter our world is still a safe abode. Nonetheless, this equality still has a life for puzzle solvers. By the application of a familiar proof, the addition of a single well-known mathematical symbol,and without resorting to prohibited mathematical operations, 1=2 can still be shown to be true. Have fun, and please use spoilers. Share this post Link to post Share on other sites

0 vistaptb 2 Report post Posted January 21, 2013 Put a slash through the equal sign? Share this post Link to post Share on other sites

0 phil1882 13 Report post Posted January 21, 2013 the most familiar proof i can think of is a^2 +b^2 = c^2. if i take the logarithm of both sides; i get log(a^2 +b^2) = log(c^2) log(a^2 +b^2) = 2*logĀ© log(a^2 +b^2)/logĀ© = 2 how to show this is also equal to 1 however has me stymied. Share this post Link to post Share on other sites

0 superprismatic 11 Report post Posted January 21, 2013 We will prove a more general result using a simple induction argument. We prove: Any set of numbers are all equal. Proof: (Base Case) -- If we have a set consisting of one number, it is clear that all the numbers in this set are equal. (Inductive Step) -- We assume that any set of N numbers are equal. Now, we will show that any set of N+1 numbers are all equal. Let us take one number out of the set of N+1. We are now left with a set of N numbers and, by our induction hypothesis, all N of these are equal. So, we put the number we took out (call this number X) back in the set to again make it a set of N+1 numbers. We now take a different number out of this set (call this number Y). Now, we once again have a set of N numbers which, by the induction hypothesis, have all equal numbers. The only way this can happen is if X=Y because each of X and Y are equal to the other numbers in the set. So, any set of N+1 numbers must all be equal. (Therefore) By induction, any set of numbers must contain numbers which are all equal. In particular, the set {17.3,1,-6,2} must contain numbers which are all equal. Thus 1=2. Share this post Link to post Share on other sites

0 bushindo 14 Report post Posted January 22, 2013 We will prove a more general result using a simple induction argument.We prove: Any set of numbers are all equal.Proof:(Base Case) -- If we have a set consisting of one number, it is clearthat all the numbers in this set are equal.(Inductive Step) -- We assume that any set of N numbers are equal.Now, we will show that any set of N+1 numbers are all equal. Let ustake one number out of the set of N+1. We are now left with a setof N numbers and, by our induction hypothesis, all N of these areequal. So, we put the number we took out (call this number X)back in the set to again make it a set of N+1 numbers. We nowtake a different number out of this set (call this number Y).Now, we once again have a set of N numbers which, by the inductionhypothesis, have all equal numbers. The only way this can happenis if X=Y because each of X and Y are equal to the other numbers inthe set. So, any set of N+1 numbers must all be equal.(Therefore) By induction, any set of numbers must contain numberswhich are all equal.In particular, the set {1,2} must contain numbers which are allequal. Thus 1=2.Counter example to this proof =)We use the same set- {1, 2}. Following the reasoning above, thenX = 1Y = 2but it doesn't follow that X = Y. I guess N=2 is the weakest link of this induction chain. Share this post Link to post Share on other sites

0 superprismatic 11 Report post Posted January 22, 2013 We will prove a more general result using a simple induction argument. We prove: Any set of numbers are all equal. Proof: (Base Case) -- If we have a set consisting of one number, it is clear that all the numbers in this set are equal. (Inductive Step) -- We assume that any set of N numbers are equal. Now, we will show that any set of N+1 numbers are all equal. Let us take one number out of the set of N+1. We are now left with a set of N numbers and, by our induction hypothesis, all N of these are equal. So, we put the number we took out (call this number X) back in the set to again make it a set of N+1 numbers. We now take a different number out of this set (call this number Y). Now, we once again have a set of N numbers which, by the induction hypothesis, have all equal numbers. The only way this can happen is if X=Y because each of X and Y are equal to the other numbers in the set. So, any set of N+1 numbers must all be equal. (Therefore) By induction, any set of numbers must contain numbers which are all equal. In particular, the set {1,2} must contain numbers which are all equal. Thus 1=2. Counter example to this proof =) We use the same set- {1, 2}. Following the reasoning above, then X = 1 Y = 2 but it doesn't follow that X = Y. I guess N=2 is the weakest link of this induction chain. it's the ONLY weak link....Give a guy a little break here! Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 22, 2013 Good thinking so far, but it's not as complicated as it looks. You'll just have to go a bit more outside the box. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 23, 2013 Put a slash through the equal sign? Actually not bad. But not the solution I had in mind. Share this post Link to post Share on other sites

0 superprismatic 11 Report post Posted January 23, 2013 Let T be the square root of 2. Consider the infinite continued exponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearly have X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4 because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets 1=2. QED Share this post Link to post Share on other sites

0 superprismatic 11 Report post Posted January 23, 2013 Start with the trig identity cos(2x)=cos^{2}(x)-sin^{2}(x) re-arrange the equation to cos(2x)+sin^{2}(x)=cos^{2}(x) take square roots sqrt(cos(2x)+sin^{2}(x))=cos(x) divide by sqrt(2) to get sqrt(cos(2x)+sin^{2}(x))/sqrt(2)=cos(x)/sqrt(2) add (3/2) to both sides (3/2)+sqrt(cos(2x)+sin^{2}(x))/sqrt(2)=(3/2)+cos(x)/sqrt(2) evaluate at x=(3*pi/4) to get 2=1. Voila! Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 23, 2013 Let T be the square root of 2. Consider the infinite continuedexponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearlyhave X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets1=2. QED This is great - beautiful, even. Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.) I made a graph of the divergence point sometime back, I will look for it. When the records are written, however, this def gets Honorable Mention. Share this post Link to post Share on other sites

0 bushindo 14 Report post Posted January 23, 2013 Let T be the square root of 2. Consider the infinite continuedexponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearlyhave X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets1=2. QEDThis is great - beautiful, even.Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.)I made a graph of the divergence point sometime back, I will look for it.When the records are written, however, this def gets Honorable Mention. Great proof as well. I have no beef with the power tower, but isn't the argument similar to this oneLet's say that a variable x satisfies the following relationship,x^2 - 6x + 8 = 0Since x=2 and x=4 both satisfy the equation, then 2 = 4, or 1=2. It should be easy to find the weak point now. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 23, 2013 Start with the trig identitycos(2x)=cos^{2}(x)-sin^{2}(x)re-arrange the equation tocos(2x)+sin^{2}(x)=cos^{2}(x)take square rootssqrt(cos(2x)+sin^{2}(x))=cos(x)divide by sqrt(2) to getsqrt(cos(2x)+sin^{2}(x))/sqrt(2)=cos(x)/sqrt(2)add (3/2) to both sides(3/2)+sqrt(cos(2x)+sin^{2}(x))/sqrt(2)=(3/2)+cos(x)/sqrt(2)evaluate at x=(3*pi/4) to get2=1.Voila!Ah, the old dazzle 'em with double angles and parentheses ploy! Remember it well from my college days. Did you think I would actually evaluate all those functions? Unfortunately that is unnecessary. It came out wrong by a factor of 4. While the problem asked for a "proof" that 1=2; you ended up with 2=1; I.e., that 4=2. In a dyslexic world, where a skeptic is unsure there is such a thing as a dog, that might suffice. But I'm afraid the quest must go on. Efforts should be rewarded, nonetheless. Thus, a clue. If you get far enough out of the box, it might be the start of a relaxing hour after work. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 23, 2013 Let T be the square root of 2. Consider the infinite continuedexponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearlyhave X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets1=2. QEDThis is great - beautiful, even.Except, the power tower unfortunately does not converge for both cases (for 2, but alas not for 4 as I remember.)I made a graph of the divergence point sometime back, I will look for it.When the records are written, however, this def gets Honorable Mention. Great proof as well. I have no beef with the power tower, but isn't the argument similar to this oneLet's say that a variable x satisfies the following relationship,x^2 - 6x + 8 = 0Since x=2 and x=4 both satisfy the equation, then 2 = 4, or 1=2. It should be easy to find the weak point now.Nicely debunked Bushindo. However, as I stand now the the company of two great logicians/mathematicians, I tread lightly regarding weak points, knowing the desired solution might likewise be so described. Its redeeming quality being only that it fits the given conditions so nicely. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 25, 2013 Write: For x > 0, x + x + x + ... + x (x terms) = x^{2} Differentiate with respect to x: 1 + 1 + 1 + ... + 1 (x terms) = 2x x = 2x. Divide by x > 0. 1 = 2. No. Sorry that's not it, either. Keep trying. Share this post Link to post Share on other sites

0 BobbyGo 7 Report post Posted January 25, 2013 I remember reading somewhere (it might have even been this site, for all I know) that 1 = 2 was used in some kind of physics conversion. Ohms to amps maybe? Or something close-about? Share this post Link to post Share on other sites

0 CaptainEd 19 Report post Posted January 25, 2013 Yes, but that's physics. My father was a nuclear physicist in the 50's, in the slide rule era. When I was in high school, he explained to me a couple of important principles of applied mathematics: * For large enough N, anything = 1 * For really large enough N, anything = 0 As a corollary, he pointed out (for a different problem than this one), that "2 is one of the larger numbers that is equal to 1". But that's physics... Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 26, 2013 Uhhh ... There! I just cut a king-size hole in the box so we can all leave.. Since we're looking for a proof and a symbol, I considered saying what the symbol is. Suffice it that the symbol would probably bring the search to an end, and specifying the proof most certainly would. So I'll just let that information itself be the clue. Relax, and let the solution come to you. Share this post Link to post Share on other sites

0 Prime 15 Report post Posted January 26, 2013 I haven't read the entire thread, not sure if someone has not suggested that already. (1 - 1.5)^{2} = 0.25 sqrt((1 - 1.5)^{2})^{}= sqrt(0.25) 1 - 1.5 = 0.5 1 = 2 Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted January 26, 2013 1 unit = 2 units vice versa Time dilation: Special Relativity A and B are reference frames moving at v m/s. So that: A observed (compares) B clock at 1 second mark while his clock is at 2 seconds mark. Same is true to: B observed (compares) A clock at 1 second mark while his clock is at 2 seconds mark. Physics are describe by mathematical means Like Lorentz transformations. Share this post Link to post Share on other sites

0 plasmid 39 Report post Posted January 26, 2013 If we're looking for a proof and a mathematical symbolthe proof is a pitcher of 10 proof lagerthe symbol is ā (infinity if your font doesn't match mine)and with beer goggles, 1 can definitely be 2 Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 27, 2013 If we're looking for a proof and a mathematical symbolthe proof is a pitcher of 10 proof lager the symbol is ā (infinity if your font doesn't match mine) and with beer goggles, 1 can definitely be 2 Thanks plasmid for taking us out of the box. With the pitcher and goggles, and infinity tho, I lost track of where the equation fit in. You might be on the right track, but we're not quite there yet. Share this post Link to post Share on other sites

0 James33 1 Report post Posted January 30, 2013 How about considering it all modulo 1? Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted January 30, 2013 my OPs give all the clues necessary to solve. Let me point out two for this case. Note that: The OP contrasts clever math - like which I think is very clever - with the answer needed to solve this puzzle. So the answer is not of that type.It asserts the equation can stand, with the addition of two things. Could it be as simple as saying 1 ten-dollar bill = 2 five-dollar bills? That would be adding two things, but those two things are so similarit really would not make a puzzle. The "aha" moment should be forthcoming shortly. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted February 15, 2013 Add a percent sign. Share this post Link to post Share on other sites

0 bonanova 78 Report post Posted February 25, 2013 Add a percent sign. 1% = 2 proof Share this post Link to post Share on other sites

It is a beginner-level bit of mathematical magic to prove that 1=2.

This demonstration probably occurs multiple times in the history of this forum.

Its fallacy depends on embedding, and cleverly hiding somewhere in the "analysis," a division by zero.

Which of course is not permitted. So regarding this matter our world is still a safe abode.

Nonetheless, this equality still has a life for puzzle solvers.

By the application of a familiar proof, the addition of a single well-known mathematical symbol,

and without resorting to prohibited mathematical operations, 1=2 can still be shown to be true.

Have fun, and please use spoilers.

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